How many formula units (particles of AgNO3) are in 5.50 grams of AgNO3?
a 1.93 x 1022 molecules
b 8.67 x 1021 molecules
C 0.032 molecules
d 6.14 1022 molecules
e, 3.22 x 1022 molecules

2.0 X 10²² molecules of AgNO₃

Explanation:

Given 5.50 grams of AgNO₃, how many formula units of AgNO₃ are contained in the 5.50 grams of AgNO₃? (formula wt. of AgNO₃ = 169,87 g/mole).

Solution:

1st convert grams to moles => moles AgNO₃ = 5.50 g AgNO₃ / 169.841g AgNO₃ = 0.032377 mole AgNO₃ 2nd calculate number of particles in 0.032377 mole AgNO₃ .      Number AgNO₃ molecules in 0.032377 mole AgNO₃ = 0.032377 mole AgNO₃ X 6.023 x 10²³ molecules of AgNO₃ /1.0 mole AgNO₃ =  1.95007 X 10²² molecules of AgNO₃ in 5.50 grams of AgNO₃. The appropriate form of the answer should contain 2 Sig,Figs. based on the data point having the least number of sig.figs. in the given data. This then is 5.50 grams of AgNO₃ which has 2 sig.figs. Therefore, the number of AgNO₃ molecules in 5.5 grams of AgNO₃ = 2.0 X 10²² molecules of AgNO₃. (Note: 1.95 rounds to 2.0).

the catch is that the top numbers are atomic masses the bottom numbers are the numbers of protons and electrons in an uncharged atom. they are the number of that atoms as you would find them on the periodic table. further, the numbers on the left are used to determine the missing numbers on the right

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer